Context-free A derivation tree or parse tree in G Grammars and - - PowerPoint PPT Presentation
Context-free A derivation tree or parse tree in G Grammars and Languages Example strings in L(G) Another string in L(G) (start or sentence) (rules) 1 2 n e o l p i t m a (only; produces or may be
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Example strings in L(G) (“start” or “sentence”) (“rules”) Another string in L(G) A derivation tree or parse tree in G
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i.e., “context-free” (only; “produces” or “may be rewritten as”)
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(reflexive, transitive closure
k≥0
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We’ll see later that Ltwo={ww|w ∈Σ*} is not context free. At first glance, you might think that adding a new start symbol S’ and a rule S’→SS to G2 would generate Ltwo, but it doesn’t; it generates all strings in Ltwo plus many others, since derivations from the two S’s are not coordinated. (Why not? It’s context-free; what happens to
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3 derivations correspond to same tree (same rules being used in the same places, just written in different orders in the linear derivation) 1) E => P+E => a+E => a+P => a+a 2) E => P+E => P+P => a+P => a+a 3) E => P+E => P+P => P+a => a+a But only one leftmost derivation corresponds to it (and vice versa). (more in HW?)
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Another grammar for the same language: E → E+E | E*E | (E) | a This grammar is ambiguous: there is a string in L(G5) with two different parse trees, or, equivalently, with 2 different leftmost
in general, (a+a)*a != a+(a*a); which is “right”? Fig 2.6: Two parse trees for a+a!a in grammar G5
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Fig 2.6: Two parse trees for a+a!a in grammar G5
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This grammar is unambiguous.
(Why? Very informally, the 3 E rules generate P((‘+’∪’*’)P)* and only via a parse tree that “hangs to the right”, as shown.)
But it has another undesirable feature: Parse tree structure does not reflect the usual precedence of * over +. E.g., tree at lower right suggests “a * a + a == a * (a + a)”
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A more complex grammar, again the same language. This one is unambiguous and its parse trees reflect usual precedence/associativity of plus and times.
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G is ambiguous L is inherently ambiguous, meaning every G for L is ambiguous
Can we always tweak the grammar to make it unambiguous? No! Language L is a CFL; grammar at left. Easy to see this G is ambiguous–strings of the form anbncn can come from the i=j (AC) or j=k (DB)
this L is also ambiguous. Intuitively, a grammar can only match a’s & b’s or b’s & c’s, not both. As a related point, { anbncn | n>0 } is not CFL.
G
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The set of context-free languages is closed under ∪, •, and *
All regular languages are CFLs Proof Sketch: Directly give simple CFLs for ∅, {ε}, and {a} for each a ∈ Σ. Combine them using the above theorem.
(Aside: We’ll later prove that CFLs are not closed under intersection or complementation.)
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G1 G2 G G G
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G G G
Then, for some x, y ε Σ* A key issue in this direction of the proof is that, since V1 ∩ V2 = ∅, there is no “crosstalk” between the two sub-grammars: any derivation in G from S1 is also a derivation in G1, and likewise S2/G2, so derivation (*) above in G can be split into (**) in G1 & G2.
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